Explanation Given: A steel tape of 150 inches used to reinforce the shipping box in all three direction with 6 inches of waste. Result used: “Let the partial derivatives f x x , f y y and f x y of the function z = f ( x , y ) exists in the circular region contained in the xy -plane with center ( a , b ) and f x ( a , b ) = 0 , f y ( a , b ) = 0 , then (a) The function has relative maximum at ( a , b ) if f x x ( a , b ) ⋅ f y y ( a , b ) − [ f x y ( a , b ) ] 2 > 0 and f x x ( a , b ) < 0 . (b) The function has relative minimum at ( a , b ) if f x x ( a , b ) ⋅ f y y ( a , b ) − [ f x y ( a , b ) ] 2 > 0 and f x x ( a , b ) > 0 . (c) If f x x ( a , b ) ⋅ f y y ( a , b ) − [ f x y ( a , b ) ] 2 < 0 , then f ( a , b ) is a saddle point (neither a maximum nor a minimum) (d) If f x x ( a , b ) ⋅ f y y ( a , b ) − [ f x y ( a , b ) ] 2 = 0 , then the test fails.” Calculation: Let x , y and z be the length, width and height of the shipping box, respectively. Then the sum of the length of steel bands in all three direction is 2 ( x + y ) + 2 ( x + z ) + 2 ( y + z ) . Also it is given that, the sum of the length of steel bands used to rein forced the box is 150 inches. Here, the length of over lapped steel bands is 6 inches Therefore, the sum of the length of steel bands used to rein forced the box without over lapped is 144 inches. That is, 2 ( x + y ) + 2 ( x + z ) + 2 ( y + z ) = 150 − 6 . Simplify the above equation, 2 ( x + y + x + z + y + z ) = 144 2 ( 2 x + 2 y + 2 z ) = 144 4 ( x + y + z ) = 144 x + y + z = 36 (1) The volume of the shipping box is equal to product of length and breadth and height. That is, V ( x , y , z ) = x y z . …… (2) From equation (1), z = 36 − x − y . Change the function V ( x , y , z ) into two variable by substituting the above z value in equation (2). V ( x , y ) = x y ( 36 − x − y ) = 36 x y − x 2 y − x y 2 Calculate the partial derivative of the above function with respect to x . V x ( x , y ) = ∂ ∂ x ( 36 x y − x 2 y − x y 2 ) = ∂ ∂ x ( 36 x y ) − ∂ ∂ x ( x 2 y ) − ∂ ∂ x ( x y 2 ) = 36 y − 2 x y − y 2 Again, take the partial derivative of the function V x ( x , y ) with respect to x . V x x ( x , y ) = ∂ ∂ x ( V x ( x , y ) ) = ∂ ∂ x ( 36 y − 2 x y − y 2 ) = ∂ ∂ x ( 36 y ) − ∂ ∂ x ( 2 x y ) − ∂ ∂ x ( y 2 ) = − 2 y That is, V x x ( x , y ) = − 2 y . …… (3) Calculate the partial derivative of the function V ( x , y ) with respect to y . V y ( x , y ) = ∂ ∂ y ( 36 x y − x 2 y − x y 2 ) = ∂ ∂ y ( 36 x y ) − ∂ ∂ y ( x 2 y ) − ∂ ∂ y ( x y 2 ) = 36 x − x 2 − 2 x y Again, take the partial derivative of the function V y ( x , y ) with respect to y . V y y ( x , y ) = ∂ ∂ y ( V y ( x , y ) ) = ∂ ∂ y ( 36 x − x 2 − 2 x y ) = ∂ ∂ y ( 36 x ) − ∂ ∂ y ( x 2 ) − ∂ ∂ y ( 2 x y ) = − 2 x That is, V y y ( x , y ) = − 2 x . …… (4) Calculate the partial derivative of the function V x ( x , y ) with respect to y